Transformation of axes Questions in English

(B) Reflection of point $P(1,4)$ about the line $y=x$ is $A(4,1)$.
After translation of point $A(4,1)$ through a distance of $1$ unit along the positive direction of $X$-axis,the new coordinate is $B(5,1)$.
After rotation of the point $B(5,1)$ through an angle $\theta = \frac{\pi}{4}$ about the origin in the anti-clockwise direction,the new coordinates $(x', y')$ are given by:
$x' = x \cos \theta - y \sin \theta = 5 \cos \frac{\pi}{4} - 1 \sin \frac{\pi}{4} = \frac{5}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
$y' = x \sin \theta + y \cos \theta = 5 \sin \frac{\pi}{4} + 1 \cos \frac{\pi}{4} = \frac{5}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.
Thus,the coordinates of $C$ are $(2\sqrt{2}, 3\sqrt{2})$.
Hence,option $B$ is correct.

(D) The transformation equations for rotating the axes by an angle $\alpha$ are given by:
$x' = x \cos \alpha + y \sin \alpha$
$y' = -x \sin \alpha + y \cos \alpha$
Given $(x, y) = (1, 2)$ and $(x', y') = \left(\frac{3 \sqrt{3}-1}{2 \sqrt{2}}, \frac{\sqrt{3}+3}{2 \sqrt{2}}\right)$:
$\frac{3 \sqrt{3}-1}{2 \sqrt{2}} = 1 \cos \alpha + 2 \sin \alpha$ $(1)$
$\frac{\sqrt{3}+3}{2 \sqrt{2}} = -1 \sin \alpha + 2 \cos \alpha$ $(2)$
Multiply $(1)$ by $2$ and $(2)$ by $1$ and add them:
$2 \left(\frac{3 \sqrt{3}-1}{2 \sqrt{2}}\right) + \frac{\sqrt{3}+3}{2 \sqrt{2}} = 2 \cos \alpha + 4 \sin \alpha - \sin \alpha + 2 \cos \alpha$
$\frac{6 \sqrt{3}-2 + \sqrt{3}+3}{2 \sqrt{2}} = 4 \cos \alpha + 3 \sin \alpha$ (This approach is complex,let's use $x'^2 + y'^2 = x^2 + y^2$)
$x'^2 + y'^2 = 1^2 + 2^2 = 5$
$\left(\frac{3 \sqrt{3}-1}{2 \sqrt{2}}\right)^2 + \left(\frac{\sqrt{3}+3}{2 \sqrt{2}}\right)^2 = \frac{27+1-6 \sqrt{3}}{8} + \frac{3+9+6 \sqrt{3}}{8} = \frac{28+12}{8} = 5$ (Consistent)
From $(1)$ and $(2)$,solving for $\alpha$ gives $\alpha = 15^{\circ} = \frac{\pi}{12}$.

(B) Step $1$: Reflection of the point $P(3,2)$ about the line $y=x$ swaps the coordinates,resulting in $(2,3)$.
Step $2$: Translation of $(2,3)$ by $3$ units in the positive direction of the $X$-axis adds $3$ to the $x$-coordinate,resulting in $(2+3, 3) = (5,3)$.
Step $3$: Rotation of the point $(x,y) = (5,3)$ through an angle $\theta = \frac{\pi}{4}$ about the origin in the counter-clockwise direction is given by the transformation:
$x' = x \cos \theta - y \sin \theta$
$y' = x \sin \theta + y \cos \theta$
Substituting $x=5, y=3, \theta = \frac{\pi}{4}$:
$x' = 5 \left(\frac{1}{\sqrt{2}}\right) - 3 \left(\frac{1}{\sqrt{2}}\right) = \frac{2}{\sqrt{2}} = \sqrt{2}$
$y' = 5 \left(\frac{1}{\sqrt{2}}\right) + 3 \left(\frac{1}{\sqrt{2}}\right) = \frac{8}{\sqrt{2}} = 4\sqrt{2}$
Thus,the final position is $(\sqrt{2}, 4\sqrt{2})$. Note: The provided option $B$ in the source was $(4\sqrt{2}, -\sqrt{2})$,which corresponds to a clockwise rotation. Given the standard counter-clockwise rotation formula,the result is $(\sqrt{2}, 4\sqrt{2})$.

(C) Let the new coordinates of the point be $(X, Y)$.
Given that the origin is shifted to $(h, k) = (-\sqrt{2}, \sqrt{2})$ and the axes are rotated anti-clockwise by an angle $\theta = 45^{\circ} = \frac{\pi}{4}$.
The original coordinates are $(x, y) = (1, -1)$.
The transformation formulas for new coordinates $(X, Y)$ are:
$X = (x - h) \cos \theta + (y - k) \sin \theta$
$Y = -(x - h) \sin \theta + (y - k) \cos \theta$
Substituting the values:
$X = (1 - (-\sqrt{2})) \cos(45^{\circ}) + (-1 - \sqrt{2}) \sin(45^{\circ})$
$X = (1 + \sqrt{2}) \frac{1}{\sqrt{2}} + (-1 - \sqrt{2}) \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} - 1 = 0$
$Y = -(1 + \sqrt{2}) \sin(45^{\circ}) + (-1 - \sqrt{2}) \cos(45^{\circ})$
$Y = -(1 + \sqrt{2}) \frac{1}{\sqrt{2}} + (-1 - \sqrt{2}) \frac{1}{\sqrt{2}}$
$Y = -\frac{1}{\sqrt{2}} - 1 - \frac{1}{\sqrt{2}} - 1 = -2 - \frac{2}{\sqrt{2}} = -2 - \sqrt{2}$
Thus,the new coordinates are $(0, -2-\sqrt{2})$.

(A) $1$. Reflection of $P(1,3)$ about $y=x$ gives $Q(3,1)$.
$2$. Translation of $Q(3,1)$ by $3$ units along the positive $X$-axis gives $R(3+3, 1) = R(6,1)$.
$3$. Rotation of $R(6,1)$ by $\theta = -\frac{\pi}{6}$ (clockwise) about the origin:
$x' = x \cos(\theta) - y \sin(\theta) = 6 \cos(-\frac{\pi}{6}) - 1 \sin(-\frac{\pi}{6}) = 6(\frac{\sqrt{3}}{2}) - 1(-\frac{1}{2}) = \frac{6\sqrt{3}+1}{2}$
$y' = x \sin(\theta) + y \cos(\theta) = 6 \sin(-\frac{\pi}{6}) + 1 \cos(-\frac{\pi}{6}) = 6(-\frac{1}{2}) + 1(\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}-6}{2}$
The final position is $\left(\frac{6 \sqrt{3}+1}{2}, \frac{\sqrt{3}-6}{2}\right)$.

(B) Since the axes are rotated through an angle $\theta = 45^{\circ}$,we replace $(x, y)$ with $(x \cos 45^{\circ} - y \sin 45^{\circ}, x \sin 45^{\circ} + y \cos 45^{\circ})$,which is $\left(\frac{x-y}{\sqrt{2}}, \frac{x+y}{\sqrt{2}}\right)$.
Substituting these into the equation $3x^2 + 3y^2 + 2xy = 2$:
$3\left(\frac{x-y}{\sqrt{2}}\right)^2 + 3\left(\frac{x+y}{\sqrt{2}}\right)^2 + 2\left(\frac{x-y}{\sqrt{2}}\right)\left(\frac{x+y}{\sqrt{2}}\right) = 2$
$\frac{3}{2}(x^2 + y^2 - 2xy) + \frac{3}{2}(x^2 + y^2 + 2xy) + \frac{2}{2}(x^2 - y^2) = 2$
$\frac{3}{2}(2x^2 + 2y^2) + (x^2 - y^2) = 2$
$3x^2 + 3y^2 + x^2 - y^2 = 2$
$4x^2 + 2y^2 = 2$
Dividing by $2$,we get $2x^2 + y^2 = 1$.

(A) Let the old coordinates be $(x, y)$ and the new coordinates be $(x', y')$. The transformation equations for rotation of axes by an angle $\theta$ are:
$x = x' \cos \theta - y' \sin \theta$
$y = x' \sin \theta + y' \cos \theta$
Given $\theta = 45^{\circ}$,$x' = \sqrt{2}$,and $y' = 4$.
Substituting these values:
$x = \sqrt{2} \cos 45^{\circ} - 4 \sin 45^{\circ} = \sqrt{2} \left( \frac{1}{\sqrt{2}} \right) - 4 \left( \frac{1}{\sqrt{2}} \right) = 1 - \frac{4}{\sqrt{2}} = 1 - 2\sqrt{2}$
$y = \sqrt{2} \sin 45^{\circ} + 4 \cos 45^{\circ} = \sqrt{2} \left( \frac{1}{\sqrt{2}} \right) + 4 \left( \frac{1}{\sqrt{2}} \right) = 1 + \frac{4}{\sqrt{2}} = 1 + 2\sqrt{2}$
Thus,the coordinates in the old system are $(1 - 2\sqrt{2}, 1 + 2\sqrt{2})$.

(A) The given equation is $3x^2+4xy+y^2-8x-4y-4=0$.
To eliminate the linear terms,we shift the origin to $(h, k)$.
Let $x = X+h$ and $y = Y+k$. Substituting these into the equation:
$3(X+h)^2 + 4(X+h)(Y+k) + (Y+k)^2 - 8(X+h) - 4(Y+k) - 4 = 0$.
Expanding this,we get:
$3X^2 + 4XY + Y^2 + X(6h+4k-8) + Y(4h+2k-4) + (3h^2+4hk+k^2-8h-4k-4) = 0$.
For the linear terms to vanish,we set the coefficients of $X$ and $Y$ to zero:
$6h+4k-8 = 0 \Rightarrow 3h+2k=4$
$4h+2k-4 = 0 \Rightarrow 2h+k=2$
Solving these,we get $h=0$ and $k=2$.
Substituting $h=0, k=2$ into the constant term:
$c = 3(0)^2 + 4(0)(2) + (2)^2 - 8(0) - 4(2) - 4 = 4 - 8 - 4 = -8$.
Thus,the transformed equation is $f(X, Y) = 3X^2 + 4XY + Y^2 - 8 = 0$.
Therefore,$f(1,1) = 3(1)^2 + 4(1)(1) + (1)^2 - 8 = 3 + 4 + 1 - 8 = 0$.

(B) The equation of the line with intercepts $a=5$ and $b=7$ is $\frac{x}{5} + \frac{y}{7} = 1$,which simplifies to $7x + 5y = 35$.
When the axes are rotated by an angle $\theta$,the new coordinates $(x', y')$ are given by $x = x' \cos \theta - y' \sin \theta$ and $y = x' \sin \theta + y' \cos \theta$.
Substituting these into the line equation: $7(x' \cos \theta - y' \sin \theta) + 5(x' \sin \theta + y' \cos \theta) = 35$.
Rearranging terms: $x'(7 \cos \theta + 5 \sin \theta) + y'(5 \cos \theta - 7 \sin \theta) = 35$.
The intercepts on the new axes are $a' = \frac{35}{7 \cos \theta + 5 \sin \theta}$ and $b' = \frac{35}{5 \cos \theta - 7 \sin \theta}$.
Since the intercepts are equal,$a' = b'$,so $7 \cos \theta + 5 \sin \theta = 5 \cos \theta - 7 \sin \theta$.
$12 \sin \theta = -2 \cos \theta$.
$\tan \theta = -\frac{2}{12} = -\frac{1}{6}$.
Therefore,$|\tan \theta| = \frac{1}{6}$.

(B) Step $1$: Shifting the origin to $(h, k)$ to remove first-degree terms from $2x^2 - 3xy + 4y^2 + 5y - 6 = 0$. Let $x = X + h$ and $y = Y + k$. Substituting these into the equation,the coefficients of $X$ and $Y$ must be zero.
$4h - 3k = 0$ and $-3h + 8k + 5 = 0$.
Solving these,we get $h = -\frac{15}{23}$ and $k = -\frac{20}{23}$.
The origin is shifted to $(h, k)$,so the point $(a, b)$ is $(-\frac{15}{23}, -\frac{20}{23})$.
Step $2$: Substitute $a = -\frac{15}{23}$ and $b = -\frac{20}{23}$ into the equation $ax^2 + 23abxy + by^2 = 0$.
$-\frac{15}{23}x^2 + 23(-\frac{15}{23})(-\frac{20}{23})xy - \frac{20}{23}y^2 = 0$.
Multiplying by $-23$,we get $15x^2 - 300xy + 20y^2 = 0$,or $3x^2 - 60xy + 4y^2 = 0$.
Step $3$: To remove the $xy$-term by rotating axes by angle $\theta$,the formula is $\tan 2\theta = \frac{B}{A - C}$,where the equation is $Ax^2 + Bxy + Cy^2 = 0$.
Here $A = 3, B = -60, C = 4$.
$\tan 2\theta = \frac{-60}{3 - 4} = \frac{-60}{-1} = 60$.

(B) To eliminate the $xy$ term in a general quadratic equation $ax^2 + bxy + cy^2 + dx + ey + f = 0$,the axes must be rotated by an angle $\theta$ such that $\tan 2\theta = \frac{b}{a-c}$.
For $A_1$: $a=3, b=5, c=3$. $\tan 2\theta_1 = \frac{5}{3-3} = \infty$ $\Rightarrow 2\theta_1 = \frac{\pi}{2}$ $\Rightarrow \theta_1 = \frac{\pi}{4}$.
For $A_2$: $a=5, b=2\sqrt{3}, c=3$. $\tan 2\theta_2 = \frac{2\sqrt{3}}{5-3} = \sqrt{3}$ $\Rightarrow 2\theta_2 = \frac{\pi}{3}$ $\Rightarrow \theta_2 = \frac{\pi}{6}$.
For $A_3$: $a=4, b=\sqrt{3}, c=5$. $\tan 2\theta_3 = \frac{\sqrt{3}}{4-5} = -\sqrt{3}$ $\Rightarrow 2\theta_3 = \frac{2\pi}{3}$ $\Rightarrow \theta_3 = \frac{\pi}{3}$.
Comparing the angles: $\frac{\pi}{3} > \frac{\pi}{4} > \frac{\pi}{6}$,which means $\theta_3 > \theta_1 > \theta_2$.

(D) Given,origin is shifted to $(h, k) = (1, -2)$.
Let the original coordinates be $(x, y) = (3, -2)$.
The new coordinates $(\alpha, \beta)$ after translation are given by $\alpha = x - h$ and $\beta = y - k$.
$\alpha = 3 - 1 = 2$ and $\beta = -2 - (-2) = 0$.
So,$(\alpha, \beta) = (2, 0)$.
Now,the axes are rotated about the origin by an angle $\theta = 45^{\circ}$.
The new coordinates $(x', y')$ after rotation are given by:
$x' = \alpha \cos \theta + \beta \sin \theta = 2 \cos 45^{\circ} + 0 \sin 45^{\circ} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$.
$y' = -\alpha \sin \theta + \beta \cos \theta = -2 \sin 45^{\circ} + 0 \cos 45^{\circ} = -2 \times \frac{1}{\sqrt{2}} = -\sqrt{2}$.
Thus,the transformed coordinates are $(\sqrt{2}, -\sqrt{2})$.

(A) Given that the origin is translated from $(0, 0)$ to $(h, k) = (-1, 2)$.
We use the transformation equations $x = X + h$ and $y = Y + k$.
Substituting the values,we get $x = X - 1$ and $y = Y + 2$.
Substitute these into the original equation $2x^2 + y^2 - 3x + 5y - 8 = 0$:
$2(X - 1)^2 + (Y + 2)^2 - 3(X - 1) + 5(Y + 2) - 8 = 0$
$2(X^2 - 2X + 1) + (Y^2 + 4Y + 4) - 3X + 3 + 5Y + 10 - 8 = 0$
$2X^2 - 4X + 2 + Y^2 + 4Y + 4 - 3X + 5Y + 5 = 0$
Combining like terms: $2X^2 + Y^2 - 7X + 9Y + 11 = 0$.
Replacing $(X, Y)$ with $(x, y)$,the transformed equation is $2x^2 + y^2 - 7x + 9y + 11 = 0$.

(C) The given equation is $x^2+6xy+8y^2=10$ $\dots$ $(i)$.
Since the axes are rotated through an angle $\theta = \frac{\pi}{4}$,the transformation equations are:
$x = x_1 \cos \frac{\pi}{4} - y_1 \sin \frac{\pi}{4} = \frac{x_1-y_1}{\sqrt{2}}$
$y = x_1 \sin \frac{\pi}{4} + y_1 \cos \frac{\pi}{4} = \frac{x_1+y_1}{\sqrt{2}}$
Substituting these into $(i)$:
$\left(\frac{x_1-y_1}{\sqrt{2}}\right)^2 + 6\left(\frac{x_1-y_1}{\sqrt{2}}\right)\left(\frac{x_1+y_1}{\sqrt{2}}\right) + 8\left(\frac{x_1+y_1}{\sqrt{2}}\right)^2 = 10$
$\frac{x_1^2+y_1^2-2x_1y_1}{2} + \frac{6(x_1^2-y_1^2)}{2} + \frac{8(x_1^2+y_1^2+2x_1y_1)}{2} = 10$
Multiplying by $2$:
$(x_1^2+y_1^2-2x_1y_1) + (6x_1^2-6y_1^2) + (8x_1^2+8y_1^2+16x_1y_1) = 20$
$(1+6+8)x_1^2 + (1-6+8)y_1^2 + (-2+16)x_1y_1 = 20$
$15x_1^2 + 3y_1^2 + 14x_1y_1 = 20$
Thus,the transformed equation is $15x^2+14xy+3y^2=20$.

(D) Given equation: $S \equiv 2x^2 - xy - y^2 - 3x + 3y = 0$.
To eliminate first-degree terms,we find the new origin $(h, k)$ by solving $\frac{\partial S}{\partial x} = 0$ and $\frac{\partial S}{\partial y} = 0$.
$\frac{\partial S}{\partial x} = 4x - y - 3 = 0$ and $\frac{\partial S}{\partial y} = -x - 2y + 3 = 0$.
Solving these,we get $x = 1, y = 1$,so $(h, k) = (1, 1)$.
To eliminate the $xy$-term by rotation,we use the formula $\tan 2\theta = \frac{B}{A - C}$,where the equation is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$.
Here $A = 2, B = -1, C = -1$.
Thus,$\tan 2\theta = \frac{-1}{2 - (-1)} = \frac{-1}{3}$.
Since $h = 1$ and $k = 1$,we have $h = k$,so $\tan 2\theta = -\frac{h}{3k} = -\frac{1}{3}$.
Therefore,the correct option is $D$.

(B) Let the original equation be $f(x, y) = 2x^2+xy-6y^2-13x+9y+15=0$.
When the origin is shifted to $(a, b)$,we substitute $x = X+a$ and $y = Y+b$.
The equation becomes $2(X+a)^2 + (X+a)(Y+b) - 6(Y+b)^2 - 13(X+a) + 9(Y+b) + 15 = 0$.
Expanding this,the linear terms in $X$ and $Y$ must vanish for the equation to take the form $2X^2+XY-6Y^2+k=0$.
The partial derivatives $\frac{\partial f}{\partial x} = 4x+y-13 = 0$ and $\frac{\partial f}{\partial y} = x-12y+9 = 0$ give the center $(a, b)$.
Solving $4a+b=13$ and $a-12b=-9$,we get $a=3$ and $b=1$.
Substituting $x=X+3$ and $y=Y+1$ into the original equation:
$2(X+3)^2 + (X+3)(Y+1) - 6(Y+1)^2 - 13(X+3) + 9(Y+1) + 15 = 0$.
$2(X^2+6X+9) + (XY+X+3Y+3) - 6(Y^2+2Y+1) - 13X-39 + 9Y+9 + 15 = 0$.
$2X^2+12X+18 + XY+X+3Y+3 - 6Y^2-12Y-6 - 13X+9Y+9+15 = 0$.
Grouping terms: $2X^2+XY-6Y^2 + (12+1-13)X + (3-12+9)Y + (18+3-6-39+9+15) = 0$.
$2X^2+XY-6Y^2 + 0X + 0Y + 0 = 0$.
Thus,$k=0$.

(A) The original equation is $2x^2 - xy + y^2 + 2x + 3y + 1 = 0$.
To shift the origin to $(h, k)$,we substitute $x = X + h$ and $y = Y + k$.
The new equation is $2(X+h)^2 - (X+h)(Y+k) + (Y+k)^2 + 2(X+h) + 3(Y+k) + 1 = 0$.
Expanding this,the linear terms in $X$ and $Y$ are $(4h - k + 2)X + (-h + 2k + 3)Y = 0$.
For the origin to be the new center,these coefficients must be zero:
$4h - k = -2$ and $-h + 2k = -3$.
Solving these,we get $h = -1$ and $k = -2$.
Thus,$h + k = -3$.
The equation becomes $2X^2 - XY + Y^2 + C' = 0$.
To eliminate the $XY$ term by rotation,we use $\tan 2\theta = \frac{2H}{A-B}$,where the equation is $AX^2 + 2HXY + BY^2 = 0$.
Here $A = 2, H = -1/2, B = 1$.
So,$\tan 2\theta = \frac{2(-1/2)}{2 - 1} = \frac{-1}{1} = -1$.
Therefore,$h + k + \tan 2\theta = -3 + (-1) = -4$.

(B) The rotation of axes by an angle $\theta = 30^{\circ}$ is given by the transformation:
$x = X \cos \theta - Y \sin \theta = X \frac{\sqrt{3}}{2} - Y \frac{1}{2}$
$y = X \sin \theta + Y \cos \theta = X \frac{1}{2} + Y \frac{\sqrt{3}}{2}$
Substituting these into the equation $4x^2+12xy+9y^2+6x+9y+2=0$:
Note that $4x^2+12xy+9y^2 = (2x+3y)^2$.
Substituting $x$ and $y$:
$2x+3y = 2(\frac{\sqrt{3}}{2}X - \frac{1}{2}Y) + 3(\frac{1}{2}X + \frac{\sqrt{3}}{2}Y) = X(\sqrt{3} + \frac{3}{2}) + Y(\frac{3\sqrt{3}-2}{2})$.
Squaring this gives the $X^2, XY, Y^2$ terms.
The linear part is $6x+9y = 6(\frac{\sqrt{3}}{2}X - \frac{1}{2}Y) + 9(\frac{1}{2}X + \frac{\sqrt{3}}{2}Y) = X(3\sqrt{3} + \frac{9}{2}) + Y(\frac{9\sqrt{3}-6}{2})$.
Thus,$2g = \frac{6\sqrt{3}+9}{2}$ and $2f = \frac{9\sqrt{3}-6}{2}$.
Calculating the ratio:
$\frac{g}{f} = \frac{6\sqrt{3}+9}{9\sqrt{3}-6} = \frac{3(2\sqrt{3}+3)}{3(3\sqrt{3}-2)} = \frac{3+2\sqrt{3}}{3\sqrt{3}-2}$.
Therefore,option $B$ is correct.

(A) Comparing the given equation with $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=4, h=4, b=10, g=-4, f=-22$ and $c=14$.
To eliminate the first degree terms,the origin $(0,0)$ must be shifted to $(h_0, k_0)$ where $h_0 = \frac{bg-fh}{h^2-ab}$ and $k_0 = \frac{af-gh}{h^2-ab}$.
Calculating the denominator: $h^2-ab = 4^2 - (4)(10) = 16 - 40 = -24$.
Calculating $h_0$: $h_0 = \frac{(10)(-4) - (-22)(4)}{-24} = \frac{-40 + 88}{-24} = \frac{48}{-24} = -2$.
Calculating $k_0$: $k_0 = \frac{(4)(-22) - (-4)(4)}{-24} = \frac{-88 + 16}{-24} = \frac{-72}{-24} = 3$.
Thus,the origin must be shifted to $(-2, 3)$.

(C) Let the origin be shifted to $P(h, k)$. The transformation equations are $x = x' + h$ and $y = y' + k$.
Substituting these into the first equation $3x^2 + y^2 - 6x + 4y + 4 = 0$:
$3(x' + h)^2 + (y' + k)^2 - 6(x' + h) + 4(y' + k) + 4 = 0$
$3(x'^2 + 2hx' + h^2) + (y'^2 + 2ky' + k^2) - 6x' - 6h + 4y' + 4k + 4 = 0$
$3x'^2 + y'^2 + (6h - 6)x' + (2k + 4)y' + (3h^2 + k^2 - 6h + 4k + 4) = 0$.
To remove the first-degree terms,set the coefficients of $x'$ and $y'$ to zero:
$6h - 6 = 0 \Rightarrow h = 1$
$2k + 4 = 0 \Rightarrow k = -2$.
So,the origin is shifted to $P(1, -2)$.
Now,substitute $x = x' + 1$ and $y = y' - 2$ into the second equation $2x^2 + 3xy - 5y^2 + 2x - 23y - 24 = 0$:
$2(x' + 1)^2 + 3(x' + 1)(y' - 2) - 5(y' - 2)^2 + 2(x' + 1) - 23(y' - 2) - 24 = 0$
$2(x'^2 + 2x' + 1) + 3(x'y' - 2x' + y' - 2) - 5(y'^2 - 4y' + 4) + 2x' + 2 - 23y' + 46 - 24 = 0$
$2x'^2 + 4x' + 2 + 3x'y' - 6x' + 3y' - 6 - 5y'^2 + 20y' - 20 + 2x' - 23y' + 46 - 24 = 0$
Grouping the terms:
$2x'^2 + 3x'y' - 5y'^2 + (4 - 6 + 2)x' + (3 + 20 - 23)y' + (2 - 6 - 20 + 46 - 24) = 0$
$2x'^2 + 3x'y' - 5y'^2 + 0x' + 0y' + 0 = 0$.
Thus,the transformed equation is $2x^2 + 3xy - 5y^2 = 0$.

(B) Let the origin be shifted to $(p, q)$ such that the new axes are parallel to the original axes. The transformation equations are $x = x' + p$ and $y = y' + q$.
Substituting these into the given equation $2x^2 + 3xy + 4y^2 + x + 18y + 25 = 0$:
$2(x' + p)^2 + 3(x' + p)(y' + q) + 4(y' + q)^2 + (x' + p) + 18(y' + q) + 25 = 0$
Expanding the terms:
$2(x'^2 + 2px' + p^2) + 3(x'y' + qx' + py' + pq) + 4(y'^2 + 2qy' + q^2) + x' + p + 18y' + 18q + 25 = 0$
Grouping the terms by $x'$,$y'$,and constants:
$2x'^2 + 3x'y' + 4y'^2 + (4p + 3q + 1)x' + (3p + 8q + 18)y' + (2p^2 + 3pq + 4q^2 + p + 18q + 25) = 0$
Comparing this with the given transformed equation $2x^2 + 3xy + 4y^2 = 1$ (or $2x'^2 + 3x'y' + 4y'^2 - 1 = 0$),the coefficients of $x'$ and $y'$ must be zero:
$4p + 3q + 1 = 0$ ... $(i)$
$3p + 8q + 18 = 0$ ... (ii)
Solving these linear equations:
From $(i)$,$4p = -3q - 1 \implies p = \frac{-3q - 1}{4}$.
Substituting into (ii): $3(\frac{-3q - 1}{4}) + 8q + 18 = 0$
$-9q - 3 + 32q + 72 = 0 \implies 23q + 69 = 0 \implies q = -3$.
Substituting $q = -3$ into $(i)$: $4p + 3(-3) + 1 = 0 \implies 4p - 8 = 0 \implies p = 2$.
Thus,$p = 2, q = -3$.

(B) Let the new coordinates be $(x', y')$. The transformation equations are $x = x' + 2$ and $y = y' + 3$.
Substituting these into the given equation $x^{2} + y^{2} - 4x - 6y + 9 = 0$:
$(x' + 2)^{2} + (y' + 3)^{2} - 4(x' + 2) - 6(y' + 3) + 9 = 0$
Expanding the terms:
$(x'^{2} + 4x' + 4) + (y'^{2} + 6y' + 9) - 4x' - 8 - 6y' - 18 + 9 = 0$
Combining like terms:
$x'^{2} + y'^{2} + (4x' - 4x') + (6y' - 6y') + (4 + 9 - 8 - 18 + 9) = 0$
$x'^{2} + y'^{2} - 4 = 0$
Therefore,the new equation is $x^{2} + y^{2} = 4$.